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vizakenjack
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Okay, so why is it that Vc assumed to be 12v - 0.7v (D1), how about contribution from the 9v battery?? Why is this not affecting the voltage at node C?
Why Vc isn't 9v - 0.7v (D2) = 8.3v?
Even if no current flows, voltage would still be there, no? Infinite resistor would have a voltage across it...Baluncore said:A reverse biassed diode drops whatever voltage is available, no current flows.
As no current flows through a reverse biassed diode, it effectively disconnects part of the circuit, a bit like an infinite resistor or insulator.
I believe so, seeing as how potential difference does not actually require current to flow, since electrical fields can exist in a vacuum. Voltage just represents stored energy that could be put to work if it was released.vizakenjack said:Even if no current flows, voltage would still be there, no? Infinite resistor would have a voltage across it...
So if voltage across open (VD2) is -2.3V (9-11.3), how come it's not affecting the voltage at VcCOWilliam said:I believe so
Vc is only affected by D1 but not D2, why? There's voltage across both diodes... Even though D2 is off.Baluncore said:There is zero current but there is still a voltage gradient and field. The circuit node that the turned off diode is connected to has a very low resistance which sets the current and so sets the voltage gradient across the reverse biased diode.
The voltages are determined by the currents flowing through the components.vizakenjack said:Vc is only affected by D1 but not D2, why? There's voltage across both diodes... Even though D2 is off.
Wait a sec, so for negative clipper in post #7Baluncore said:The voltages are determined by the currents flowing through the components.
If a diode is reverse biassed, no current flows, the voltage across the diode is irrelevant and the reverse biassed diode can be removed from the circuit diagram without changing the analysis.
The minimum output voltage, Vo, will be Vref - Vd.vizakenjack said:So why is Vo = -Vd + Vref. Wouldn't Vi affect it too?
So in this simple circuit, Vo = 3v?Baluncore said:The output remains supported at Vref-Vd as the Vi component appears across Rs, not at Vo, which is controlled by Vref-Vd.
Vo must be the 3V battery voltage because that is what the output is connected to. How could it be different ?vizakenjack said:So in this simple circuit, Vo = 3v?
so if R1 = 1k ohm, IR would be = (5v-3v)/1000 = 2 mA, current across R1.Baluncore said:The battery is being charged by the 5V source through R1. Charge current = (5V - 3V) / R1
Why not.vizakenjack said:See, I thought that voltage at that point would be the same as Vo, but apparently not ...
Baluncore said:Why not.
I do not follow your logic and cannot see why you have that opinion.vizakenjack said:Well then Vo should be 5V-3V = 2V not 3V. Don't you see the contradiction?
Measured relative to what. You show Vo is clearly being measured relative to ground.vizakenjack said:Where is for Vo Potential difference is measured between one voltage source only, which is 3V...
The definition of an ideal battery is one that maintains its voltage, whatever current is taken from it. Your arrow (Vo) must be at 3V. Quite incidentally, current will flow through R1, given by 2/R1.vizakenjack said:so if R1 = 1k ohm, IR would be = (5v-3v)/1000 = 2 mA, current across R1.
Voltage across R1 is 5V-3V = 2V
But at this point here:
Isn't the voltage also 5V-3V=2V?
See, I thought that voltage at that point would be the same as Vo, but apparently not ...
A circuit with two voltage sources is a type of electrical circuit where there are two power sources providing different voltages. This can be seen in a variety of electronic devices, from simple battery-powered devices to more complex circuits with multiple power sources.
To find the voltage at nodes in a circuit with two voltage sources, you can use Kirchhoff's Voltage Law (KVL). This law states that the sum of the voltages around a closed loop in a circuit must equal zero. By using KVL and solving for the unknown voltages, you can determine the voltage at each node in the circuit.
Yes, it is possible to have two voltage sources with the same voltage in a circuit. In this case, the two sources can be considered redundant and one can be removed without affecting the overall circuit. However, it is more common for two voltage sources to have different voltages in order to power different components in the circuit.
The purpose of having multiple voltage sources in a circuit is to provide different voltages for different components. This allows for a more efficient use of power and can also help to regulate the voltage levels for specific components that require a certain voltage to function properly.
Some common challenges when working with circuits with two voltage sources include properly calculating the voltage at nodes, ensuring that the voltage sources are connected in the correct polarity, and dealing with potential differences between the two sources. It is also important to consider the overall power and current requirements of the circuit to ensure that the voltage sources can provide enough power for all components.